Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main{
	static int mod = 1000000000 + 7;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int N = sc.nextInt();
			double[] p = new double[N];
			for(int i = 0; i < N; i++) p[i] = sc.nextDouble();
			double[][] dp = new double[N + 1][N + 1];
			dp[0][0] = 1;
			for(int i = 0; i < N; i++) {
				for(int j = 0; j < N; j++) {
					if(dp[i][j] == 0) continue;
					dp[i + 1][j + 1] += dp[i][j] * p[i];
					dp[i + 1][j] += dp[i][j] * (1 - p[i]);
				}
			}
			double ans = 0;
			for(int i = (N + 1) / 2; i <= N; i++) {
				ans += dp[N][i];
			}
			System.out.println(ans);
		}
	}
}